How Does Angular Velocity Change in a Swinging Rod with Masses?

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SUMMARY

The discussion focuses on calculating the angular velocity of a massless rod with two attached masses as it swings from a horizontal position to a vertical position. The total moment of inertia is determined to be (5/4)mL², and the initial potential energy is calculated as 2mgL. At the lowest point, the potential energy of the middle mass is mgL, and the system's kinetic energy includes both rotational and translational components. The solution requires careful algebraic manipulation to equate the total energy at the top and bottom of the swing.

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Homework Statement



A massless rod length L has a small mass m attached to the center and another mass m attached at one end. On the opposite end, the rod is hinged to a frictionless hinge. The rod is released from rest at a horizontal position and swings down. What is the angular velocity as it swings through its lowest (vertical) point? Solve in terms of g and L.

Homework Equations



use moment of inertia, energy conservation

The Attempt at a Solution



I determined the total moment of inertia of the two masses to be (5/4)mL^2. I know that initial potential energy is 2mgL (setting bottom point of swing as zero). At the bottom, the mass in the middle has PE mgL, the system has KEr of (.5)Iw^2, and both masses have a translational kinetic energy. All of this must sum to the intial 2mgL? I think I am just missing a step in the algebra.
 
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The position of the mass in the middle is L/2 from one end, not L (redo your PE terms). Also, you can treat the system as being in pure rotation about the hinge.
 

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